Problem: Let $h(x)=\dfrac{1}{x^3}-\dfrac{2}{x^2}$. $h'(-1)=$
Solution: The strategy We can first rewrite each rational term of $h$ as a negative power of $x$. Then, the derivatives of these terms can be found using the power rule : $\dfrac{d}{dx}(x^n)=n\cdot x^{n-1}$ (Remember that this applies even when $n$ is negative.) Once we have $h'(x)$, we can plug $x=-1$ into it to find $h'(-1)$. Rewriting rational terms as negative powers $\begin{aligned} h(x)&=\dfrac{1}{x^3}-\dfrac{2}{x^2} \\\\ &=x^{-3}-2x^{-2} \end{aligned}$ Differentiating using the power rule $\begin{aligned} &\phantom{=}h'(x) \\\\ &=\dfrac{d}{dx}(x^{-3}-2x^{-2}) \\\\ &=\dfrac{d}{dx}(x^{-3})-2\dfrac{d}{dx}(x^{-2}) \\\\ &=-3x^{-4}-2(-2)x^{-3} \\\\ &=-3x^{-4}+4x^{-3} \\\\ &=-\dfrac{3}{x^4}+\dfrac{4}{x^3} \end{aligned}$ Evaluating $h'(x)$ So we found that $h'(x)=-\dfrac{3}{x^4}+\dfrac{4}{x^3}$. Let's plug $x=-1$ into $h'$ : $\begin{aligned} &\phantom{=}h'(-1) \\\\ &=-\dfrac{3}{(-1)^4}+\dfrac{4}{(-1)^3} \\\\ &=\dfrac{-3}{1}+\dfrac{4}{-1} \\\\ &=-7 \end{aligned}$ In conclusion, $h'(-1)=-7$.